Acid Dissociation Constants (K_a) Chemistry Tutorial

Key Concepts

• K_a, the acid dissociation constant or acid ionisation constant, is an equilibrium constant that refers to the dissociation, or ionisation, of an acid.
• For the reaction in which the acid HA dissociates to form the ions H⁺ and A^-:

  HA H⁺ + A^-

  Ka =	[H+][A-] [HA]

• K_a provides a measure of the equilibrium position

  (i) if K_a is large, the products of the dissociation reaction are favoured

  (ii) if K_a is small, undissociated acid is favoured.

• K_a provides a measure of the strength of an acid

  (i) if K_a is large, the acid is largely dissociated so the acid is strong

  (ii) if K_a is small, very little of the acid is dissociated so the acid is weak.

• The degree to which an acid dissociates, or ionizes, can be represented as a percentage:

  % dissociation (ionization) =

  [H+ at equilibrium] [acid initial]

  × 100

  (i) If %dissociation ≈ 100%, the acid is a strong acid

  (ii) If %dissociation is small, the acid is a weak acid

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Comparing Some Acid Dissociation Constants (25°C)

The value of K_a, the dissociation constant (ionisation constant), for a number of different acids at 25°C is shown below:

Compare the acid dissociation constants for methanoic acid (formic acid), HCOOH, and ethanoic acid (acetic acid), CH₃COOH:

K_a(HCOOH) = 1.8 × 10^-4 (larger K_a)
K_a(CH₃COOH) = 1.8 × 10^-5 (smaller K_a)

Methanoic acid (formic acid), HCOOH, has a larger K_a value than ethanoic acid (acetic acid) therefore it dissociates more than ethanoic acid (acetic acid).

In aqueous solutions of the same concentration under the same conditions, for example 0.1 mol L^-1 methanoic acid and 0.1 mol L^-1 ethanoic acid, there will be:

• More undissociated ethanoic acid molecules in solution than there will be undissociated methanoic acid molecules
• more hydrogen ions in the methanoic acid solution than there will be in the ethanoic acid solution.
• which means the pH of the methanoic acid will be lower than that of the ethanoic acid solution.

Learn more about the properties of carboxylic acids at carboxylic acid tutorial

Example : Calculating [H⁺], pH and %dissociation for a Strong Acid

Calculate the [H⁺], pH and %dissociation in 0.10 mol L^-1 HCl(aq) at 25°C.

1. Write the acid dissociation equation:

   HCl H⁺(aq) + Cl^-(aq)

   Because hydrochloric acid, HCl(aq), is a strong monoprotic acid, the value for its acid dissociation constant, K_a, would be extremely large.
   Infact, it is so large that we usually consider this to be a reaction that goes to completion rather than as a reaction in equilibrium:
   HCl → H⁺(aq) + Cl^-(aq)

2. Calculate the initial and equilibrium concentrations of the species present using a R.I.C.E. Table:

   HCl(aq) is a strong acid, it completely dissociates to form H⁺ and Cl^- so 0.10 moles of HCl in 1.0 L of water will completely dissociate into 0.10 moles H⁺(aq) and 0.10 moles Cl^-(aq):
   

   R.I.C.E. Table
   	reactant		products
   Reaction:	HCl	→	H+(aq)	+	Cl-(aq)
   Initial concentrations (mol L-1)	0.10		0		0
   Change in concentration (mol L-1)	-0.10		+0.10		+0.10
   Equilibrium concentrations (mol L-1)	0.10 - 0.10 = 0		0 + 0.10 = 0.10		0 + 0.10 = 0.10

   
   From the R.I.C.E. Table, at completion (equilibrium):
   [H⁺] = 0.10 mol L^-1

3. Calculate pH of this solution at equilibrium:

   pH = -log₁₀[H⁺]
   pH = -log₁₀[0.10] = 1

4. Calculate %dissociation:

   [H⁺] = 0.1 mol L^-1 (at completion)

   [HCl initial] = 0.1 mol L^-1

   %dissociation =	[H+]     [HCl initial]	× 100
   %dissociation =	0.1 0.1	× 100
   %dissociation =	100%

   % dissocation for a strong monoprotic acid is 100%.

Example : Calculating [H⁺], pH and %dissociation for a Weak Acid

Calculate the [H⁺], pH and %dissociation in 0.10 mol L^-1 HNO₂(aq).
K_a = 5.0 × 10^-4 at 25°C

1. Write the acid dissociation equation:

   HNO₂(aq) H⁺(aq) + NO₂^-(aq)

2. Write the equilibrium contstant expression for the acid dissociation:

   Ka =	[H+][NO2-] [HNO2]
   5.0 × 10-4 =	[H+][NO2-] [HNO2]

3. Calculate the initial and equilibrium concentrations of the species present using a R.I.C.E. Table:

   let x = moles of HNO₂ that dissociate to form H⁺(aq) and NO₂^-(aq)
   

   R.I.C.E. Table
   	reactant		products
   Reaction:	HNO2		H+	+	NO2-
   Initial concentrations (mol L-1)	0.10		0		0
   Change in concentrations (mol L-1)	-x		+x		+x
   Equilibrium concentrations (mol L-1)	0.10 - x		0 + x =x		0 + x = x

   Since HNO₂ is a weak acid (K_a is small), it dissociates only slightly, x will be very small compared to 0.10

   So, at equilibrium, [HNO₂] ≈ 0.10 mol L^-1

4. Substitute the equilibrium concentration values into the expression for the acid dissociation constant and solve:

   Equilibrium Law Expression:	Ka =	[H+][NO2-] [HNO2]
   Substitute known values:	5.0 × 10-4 =	[x][x] [0.10]
   Collect like terms:	5.0 × 10-4 =	x2   0.10
   Multiply both sides by 0.10	5.0 × 10-4 × 0.10 =	x2   0.10	× 0.10
   	5.0 × 10-4 × 0.10 =	x2
   	5.0 × 10-5 =	x2
   Take square root of both sides	√5.0 × 10-5 =	√x2
   	7.1 × 10-3 =	x

   From the R.I.C.E. Table above, at equilibrium
   [H⁺] = x = 7.1 × 10^-3 mol L^-1

5. Calculate pH at equilibrium:

   pH = -log₁₀[H⁺]

   pH = -log₁₀[7.1 × 10^-3] = 2.1

6. Calculate %dissociation at equilibrium:

   [H⁺] at equilibrium = 7.1 × 10^-3 mol L^-1

   [HNO₂ initial] = 0.10 mol L^-1

   %dissociation =	[H+]     [acid initial]	× 100
   %dissociation =	[7.1 × 10-3] [0.10]	× 100
   %dissociation =	7.1 %

   % dissociation for a weak acid is less than 100%, that is, the equilibrium mixture contains some undissociated molecules of acid as well as cations and anions.

Example : Calculating [OH^-] and pOH for a Weak Acid at 25°C

Calculate the [OH^-] and pOH for an aqueous solution of 0.5 mol L^-1 acetic acid (ethanoic acid).
K_a = 1.8 × 10^-5 at 25°C.

1. Write the acid dissociation equation:

   CH₃COOH(aq) H⁺(aq) + CH₃COO^-(aq)

2. Write the equilibrium law expression for the acid dissociation:

   Ka =	[H+][CH3COO-] [CH3COOH]
   1.8 × 10-5 =	[H+][CH3COO-] [CH3COOH]

3. Calculate the initial and equilibrium concentrations of the species present using a R.I.C.E. Table:

   let x = moles of CH₃COOH that dissociate to form H⁺(aq) and CH₃COO^-(aq)

   R.I.C.E. Table
   	reactant		products
   Reaction:	CH3COOH		H+	+	CH3COO-
   Initial concentrations (mol L-1)	0.5		0		0
   Change in concentrations (mol L-1)	-x		+x		+x
   Equilibrium concentrations (mol L-1)	0.5 - x		0 + x = x		0 + x = x

   Since CH₃COOH is a weak acid (K_a is small), it dissociates only slightly, x will be very small compared to 0.5

   So, at equilibrium, [CH₃COOH] ≈ 0.5 mol L^-1

4. Substitute the concentration values into the expression for the acid dissociation constant and solve:

   Equilibrium Law Expression:	Ka =	[H+][CH3COO-] [CH3COOH]
   Substitute known values:	1.8 × 10-5 =	[x][x] [0.5]
   Collect like terms:	1.8 × 10-5 =	[x]2   [0.5]
   Multiply both sides by 0.5:	1.8 × 10-5 × 0.5 =	x2   0.5	× 0.5
   	9.0 × 10-6 =	x2
   Take square root of both sides:	√9.0 × 10-6 =	√x2
   	3.0 × 10-3 =	x

   [H⁺] = x = 3.0 × 10^-3 mol L^-1

5. Calculate the concentration of OH^- in the aqueous solution of acetic acid at 25°C:

   At 25°C, K_w, the equilibrium constant for the dissociation of water, is 10^-14

   Equilibrium Law Expression:	[H+][OH-]	=	Kw
   Substitute in known values:	[3.0 × 10-3][OH-]	=	10-14
   Divide both sides by 3.0 × 10-3:	[3.0 × 10-3][OH-] 3.0 × 10-3	=	10-14     3.0 × 10-3
   	[OH-]	=	3.3 × 10-12

   [OH^-] = 3.3 × 10^-12 mol L^-1

6. Calculate pOH:

   pOH = -log₁₀[OH^-]

   pOH = -log₁₀[3.3 × 10^-12] = 11.5